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# Impossible C2 Circles Proof Question?

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1. Been on this question for about 40 minutes
Can't understand it even with the ms, can someone please explain it?
Thanks
Hasan

Attachment 537729537731
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2. Gosh, we actually did this in school today. Apparently, the diagram is part of a diagram of three overlapping circles in which a and d are the centre of each respective circles and that two equilateral triangles are made.
3. (Original post by HasanRaza1)
Been on this question for about 40 minutes
Can't understand it even with the ms, can someone please explain it?
Thanks
Hasan

Attachment 537729537731
right okay here's what you have to do....replicate the diagram first and draw in the other circles, then draw two circle sectors where the arcs of each are OC and OBThen, draw a hexagon inside the centre circle (the first one you drew)Split the hexagon into six equilateral triangles (every angle is 60 degrees) and you should see that two of the triangles fall exactly into those sectors, the sides are length r (one side of each will line up with the radius already drawn in).use Area = 1/2 absin(c) to work out the area of one of these triangles and you should get Area = 1/2 r x r sin(30) = 1/2 r^2 x (3^1/2)/2Look at the diagram again and you'll see that because of the hexagon you can tell that the two circle segments are a sixth of the circle each as each of them fits inside one of the six triangles (within the hexagon) therefore the area of ONE of the sectors is (pi x r^r)/6 If you do the area of the segment - the are of the triangle you get this small area where the arc is and if you look at the triangle that the shaded region is inside you'll see that TWO of these areas (the one you just found) cross into that triangle and the rest is the shaded region that you want.Therefore, double the area of that you just found (should be ((r)^2(3)^1/2)/4 Then do the area of one of the triangles made by the hexagon minus this area (which you just doubled) and then simplify by taking 1/6 and r^2 outside the bracket and multiplying the value with (3)^1/2 by 2 and you're sorted
4. (Original post by The Rohan)
right okay here's what you have to do....replicate the diagram first and draw in the other circles, then draw two circle sectors where the arcs of each are OC and OBThen, draw a hexagon inside the centre circle (the first one you drew)Split the hexagon into six equilateral triangles (every angle is 60 degrees) and you should see that two of the triangles fall exactly into those sectors, the sides are length r (one side of each will line up with the radius already drawn in).use Area = 1/2 absin(c) to work out the area of one of these triangles and you should get Area = 1/2 r x r sin(30) = 1/2 r^2 x (3^1/2)/2Look at the diagram again and you'll see that because of the hexagon you can tell that the two circle segments are a sixth of the circle each as each of them fits inside one of the six triangles (within the hexagon) therefore the area of ONE of the sectors is (pi x r^r)/6 If you do the area of the segment - the are of the triangle you get this small area where the arc is and if you look at the triangle that the shaded region is inside you'll see that TWO of these areas (the one you just found) cross into that triangle and the rest is the shaded region that you want.Therefore, double the area of that you just found (should be ((r)^2(3)^1/2)/4 Then do the area of one of the triangles made by the hexagon minus this area (which you just doubled) and then simplify by taking 1/6 and r^2 outside the bracket and multiplying the value with (3)^1/2 by 2 and you're sorted
Thank you so much for the help
I followed your reasoning and ended up with this?
5. (Original post by HasanRaza1)
Been on this question for about 40 minutes
Can't understand it even with the ms, can someone please explain it?
Thanks
Hasan

Attachment 537729537731
Don't know how good your integration is but assume you can do some as it's C2, don't worry if you don't follow this entirely.

We know the equation of the central circle. we also know that the unshaded area in the right half of the circle is four times the area between the circumference and the line where y=r/2.

We can then solve to find the intersection point between the top and central circles to get the intersection point

knowing the above we can set up an integral to find this area.

This integral requires a substitution to solve

giving

our upper and lower limits become and 0 respectively.

Now our integral becomes

solving this integral yields

to solve the from earlier we recall that A that we have calculated is only a quarter of the unshaded area on that side so we can multiply it by 4 and subtract from the area of a semicircle and it gives us the solution.

Again, been a while since I did A-level so not sure if that is a C2 level solution...
6. I remember doing this question, the area of the odd shaped parts either side of the shaded region is 2x (1/2 r^2 theta) - 1/2absinC, as it's two overlapping sectors, hence the area is the area of both sectors minus any overlap (the triangle bit), it's quite a tough thing to visualize, but if you understand the overlapping sectors thing you should be fine.
8. (Original post by natninja)
Don't know how good your integration is but assume you can do some as it's C2, don't worry if you don't follow this entirely.

We know the equation of the central circle. we also know that the unshaded area in the right half of the circle is four times the area between the circumference and the line where y=r/2.

We can then solve to find the intersection point between the top and central circles to get the intersection point

knowing the above we can set up an integral to find this area.

This integral requires a substitution to solve

giving

our upper and lower limits become and 0 respectively.

Now our integral becomes

solving this integral yields

to solve the from earlier we recall that A that we have calculated is only a quarter of the unshaded area on that side so we can multiply it by 4 and subtract from the area of a semicircle and it gives us the solution.

Again, been a while since I did A-level so not sure if that is a C2 level solution...
Wow I was just about to say my brain hurts reading that haha
Thank you very much anyway though I appreciate the effort
9. i don't think they expect you to use calculus for this

once you spot the equilateral triangles it is quite straightforward.
10. (Original post by HasanRaza1)
Thank you so much for the help
I followed your reasoning and ended up with this?
why have you worked out the area of the segment in the way you have?
it should just be (Pi x r^2)/6 and that should do it for you

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